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converting a DC power supply to lower voltage and current - help!

Hello - I'm interested in converting a 3.3 volt, 3 amp D.C. power
supply to 1.2 volt, 600 mAmp. Is it possible? I looked at a couple
of other posts and used some of the calcs there, but the resistor i
come up with doesn't sound right to me (3.5microOhm, 1260 Watt).
Also, if it is a resistor i'm supposed to use, does it matter which
lead it goes on (+ or -)?

Thanks in advance for the help!
Eric
 
Hello - I'm interested in converting a 3.3 volt, 3 amp D.C. power
supply to 1.2 volt, 600 mAmp. Is it possible? I looked at a couple
of other posts and used some of the calcs there, but the resistor i
come up with doesn't sound right to me (3.5microOhm, 1260 Watt).
Also, if it is a resistor i'm supposed to use, does it matter which
lead it goes on (+ or -)?

Thanks in advance for the help!
Eric

Use an LM317 adjustable voltage regulator (50 cents). Don't use a
dropping resistor to supply voltage. Your voltage will vary according
to your load.

BRW
 
J

Jonathan Kirwan

Jan 1, 1970
0
Hello - I'm interested in converting a 3.3 volt, 3 amp D.C. power
supply to 1.2 volt, 600 mAmp. Is it possible? I looked at a couple
of other posts and used some of the calcs there, but the resistor i
come up with doesn't sound right to me (3.5microOhm, 1260 Watt).
Also, if it is a resistor i'm supposed to use, does it matter which
lead it goes on (+ or -)?

Thanks in advance for the help!

One of the things you _may_ be confused about is that when there are
three variables involved, knowing two of them solves the other one
exactly. Ohm's law is like that, too: E=I*R. Three variables. If
you modify your power supply to put out 1.2 volts, that specifies one
variable, E. Whatever you plan to connect to it, the "load,"
specifies another, R. The third is then a consequence of the other
two and not something you normally design a power supply to limit. In
other words, if you modify the output voltage and hook up your load to
it, the current that flows just happens as a consequence and you don't
need to do much about that except to make sure your power supply is up
to delivering whatever is required.

Now, a lot of "loads" are dynamic, in the sense that they may try to
pull varying amounts of current over time. In these cases, you need
to make sure your modified 1.2 volt power supply is able to adapt and
keep the voltage steady over varying load currents. This is what a
power supply's usual job actually is and why people go to any trouble
at all designing one. So if your load is a varying one, like a radio
may be, then you really do need a "power supply design" of some kind
to hold that 1.2 volts over a varying R.

On the other hand, if you know your load's R very well and it doesn't
change enough to matter, then you could just compute a "dropping
resistor" value and let it go at that. In that case, the result would
look like:

,---------,
| |
(+) | R <-- dropping resistor
,----------, R
| 3.3 volt | |
| power | |
| supply | |
'----------' R <-- 1.2 volt "load"
(-) | R with known "R"
| | or known "I"
'---------'

In this case, you'd compute the dropping resistor value as (3.3 - 1.2)
divided by the load's current or else divided by (1.2 / R), where R is
the load's known resistance. So the computation is either:

(3.3 - 1.2) / I
or,
(3.3 - 1.2) / (1.2 / R)

depending on whether you know the R or the I of the load. Since you
specified 600mA, you might prefer to use the first one I mentioned.
This would work out to:

(3.3V - 1.2V) / 600mA = (3.3V - 1.2V) / 0.6A = 3.5 Ohms

Note that this corresponds vaguely to your own calculation, except
that you wrote "microOhms." That's wrong.

But be wary. It may be that your load specifies a power supply that
is ___capable___ of 600mA. This is quite different that saying that
your load actually ___needs___ 600mA, in practice. It may only
require a typical of 100mA or 200mA and they may have written down
600mA as being "safely sufficient." If that is what is really going
on, computing that single resistor value is probably NOT the way to go
and is certainly going to lead to a wrong calculation. Instead, in
that case, you probably really DO need a proper 1.2 volt power supply
that actually knows how to observe its own output and keep track of
your load's needs as they change from time to time.

In that case, just explain what you are trying to power up as best you
can (as much detail as you can muster) and ask for a specific
solution.

Jon
 
Here are the specifics of what I'm trying to do:

I've got a hair trimmer with a rechargable battery in it. I'd like to
replace the battery with a constant power supply that plugs into the
wall. The battery is 1.2V and 600mA. The closest power supply i have
is 3.3V and 3A.

It seems like the load (hair trimmer) should be fairly constant,
though maybe it changes based on whether it's cutting or not.

Anyway, thanks again for your input. If this helps clarify the best
solution, I would love to hear it!

Thanks much -
Eric
 
Here are the specifics of what I'm trying to do:

I've got a hair trimmer with a rechargable battery in it. I'd like to
replace the battery with a constant power supply that plugs into the
wall. The battery is 1.2V and 600mA. The closest power supply i have
is 3.3V and 3A.

It seems like the load (hair trimmer) should be fairly constant,
though maybe it changes based on whether it's cutting or not.

Anyway, thanks again for your input. If this helps clarify the best
solution, I would love to hear it!

Thanks much -
Eric

Well said, Jon.
Eric - Are you sure your battery says 600 mA (current), or does it say
600 mAh (capacity)? I've never seen a battery with a current rating
stamped on it, but almost all rechargable batteries have a capacity
stamped on them (mAh). A battery may have a 600 mAh capacity, but it
could possibly supply much more than 600 mA of current. You should
measure your actual current draw.

BRW
 
BRW -

I think you're right. It's probably stamped 600mAh capacity.
Unfortunately, the battery doesn't charge any longer, though i could
probably put a 1.5V AA battery in to measure the current draw, right?
Or can I go ahead and drop the voltage with the variable resistor and
see if the supply can properly regulate the current?

Eric
 
J

Jonathan Kirwan

Jan 1, 1970
0
I think you're right. It's probably stamped 600mAh capacity.
Unfortunately, the battery doesn't charge any longer, though i could
probably put a 1.5V AA battery in to measure the current draw, right?
Or can I go ahead and drop the voltage with the variable resistor and
see if the supply can properly regulate the current?

It probably won't hold the voltage, anyway. The load, if you are
using the trimmer, will change appreciably and so will the voltage
being supplied, as a consequence of that.

Some suggestions (many others exist):

(1) You might be able to find a supplier of the exact battery on the
web. I've found some cheap sources, with some effort, for some fairly
odd-ball NiCad contraptions. You might be lucky, too. This assumes
you still have some device for recharging the trimmer.

(2) Just try what you are talking about. Measure the current by
using an alkaline D cell (no reason to test this with a AA -- you want
the best voltage stability you can get cheap and easy.) Just check it
quickly, first, to make sure the extra few tenths of a volt are okay
(probably just fine.) Then go for longer and measure. Then measure
it again under some reasonable load (cut some hair.) My wild guess is
that the load current will about double, at most. Probably less than
that, though. In any case, if you plan to just use a dropping
resistor in spite of all this, you will either have too much voltage
without a load or too little voltage under load. Whether that works
out for you will depend. But you can, of course, experiment and see.

(3) Make a proper 1.2 volt power supply out of your current 3.3 volt
one or else make up a new one.

(4) Make your own custom battery pack and wire it up. Use a D-cell
NiCad and get a charger for it and add some wire to connect to the
trimmer. Or use a D-cell alkaline, if that works fine. This isn't
any more cumbersome than some modified AC-to-3.3V-to-1.2V supply and
you can get standard, cheap batteries and use standard, cheap
rechargers (if you get NiCads.)

Jon
 
J

Jon Slaughter

Jan 1, 1970
0
Hello - I'm interested in converting a 3.3 volt, 3 amp D.C. power
supply to 1.2 volt, 600 mAmp. Is it possible? I looked at a couple
of other posts and used some of the calcs there, but the resistor i
come up with doesn't sound right to me (3.5microOhm, 1260 Watt).
Also, if it is a resistor i'm supposed to use, does it matter which
lead it goes on (+ or -)?

Thanks in advance for the help!
Eric

You could use some diodes. Silicon gives 0.7V drop for each and 3 of them is
2.1V which 3.2V - 2.1V = 1.2V which is exactly what you need. (well
ultimately its a good match)

The issue here is power dissipation but if your at 600mA then thats 0.7V*1A
~ 700mW or so max. You should be able to find some diodes that can handle
this quite easily or if your worried you could run several in parallel to
split the power dissipation(although there are a few problems with this I
think it would be ok... just not to many because then you limit the current
through each diode and it might not reach your 0.7V).

What I would do if I needed a quick and dirty solution would be to but 3 1W
Si diodes in series with the supply. Then I would test it on a constant load
like a resistor(you'll need something like 20-50ohmz for a decent test) then
I would test it on the device and see how it performs. Ultimately I was
worried about ruining the device I wouldn't do this but use a regulator...
chances are though it will be ok.

Jon
 
Thanks for all of your help. This is my plan...

As I'd like to avoid rechargable batteries, i'll try the following:
I've got a corded shaver (probably a different motor) that i'll look
at to get a vague idea of the current draw (as i think you were right
about the 600mA being 600mAhours of storage). I'll then try either a
resistor or the tripple diode solution - maybe both to see which works
best.

again - thanks for your help - i'll post back once i've tried these
options.

best -
eric
 
BRW -

I think you're right. It's probably stamped 600mAh capacity.
Unfortunately, the battery doesn't charge any longer, though i could
probably put a 1.5V AA battery in to measure the current draw, right?
Or can I go ahead and drop the voltage with the variable resistor and
see if the supply can properly regulate the current?

Eric

So do what I did. Get 2 1000mA NiCad cells with solder tabs and just
replace the bad cells. I did that 3 years ago and it's going strong.
The higher capacity cells require less frequent charging as well.

GG
 
E

ehsjr

Jan 1, 1970
0
Here are the specifics of what I'm trying to do:

I've got a hair trimmer with a rechargable battery in it. I'd like to
replace the battery with a constant power supply that plugs into the
wall. The battery is 1.2V and 600mA. The closest power supply i have
is 3.3V and 3A.

It seems like the load (hair trimmer) should be fairly constant,
though maybe it changes based on whether it's cutting or not.

Anyway, thanks again for your input. If this helps clarify the best
solution, I would love to hear it!

Thanks much -
Eric

Replace the battery with a 1200 mAh rated NiCd, then
wire up this circuit to the trimmer:

-----
Vcc ---+---Vin|LM317|Vout---+
| ----- |
[.1uF] | [47R]
| | |
| +----------+----> + to trimmer
Gnd ---+-------------------------> Gnd to trimmer

That will provide a constant current to the trimmer
of ~ 26 mA. The NiCd battery will regulate the voltage
to ~ 1.2 volts. You will need to use a 5 volt or
higher supply - the closer to 5 volts the better.
Leave it plugged in all the time. The setup will
charge the battery very slowly - a bit less than
2 1/2 days for a full charge. Because it charges slowly,
it can be left plugged in all the time.

If you make the connection from the circuit to the trimmer
detachable, you'll have the best of both worlds - a cord
free trimmer when you want it that way, and a corded trimmer
when power is available.

If you can't get a battery that will fit into the trimmer,
you can use the same circuit with an external battery. You
do not want to use this circuit without a battery.

Ed
 
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